using phazors question.. - Electronic Circuits Projects Diagrams Free

transgalactic · 13th July 2009, 04:44 PM

this is the last part of a bigger question
if you i forgot to mention some data please tell

in the last part i got
$\ddot{v_c}+2\dot{v_c}+2v_c=v_s(t)$
$\dot{v_c(0)}=2$
$v_c(0)=0$

i need to find out if the response for $v_s(t)=u(t)+cos(t)$
equals the sum of $v_c^u$ (the response for spet function) and $v_c^cos$ the response for cosine
i start by finding the response for cos(t)
$v_c(t)->V_c(s)$
$\dot{v_c}->sV_c(s)-v_c(0)$
$\ddot{v_c}->s(sV_c(s)-v_c(0))-\dot{v_c(0)}$
$s(sV_c(s)-v_c(0))-\dot{v_c(0)}+2sV_c(s)-v_c(0)+2V_c(s)=\frac{s}{s^2+1}$
$ V_c(s)[s^2+2s+2]+2=\frac{s}{s^2+1} $
so i get
$ V_c(s)=\frac{s}{(s^2+2s+2)(s^2+1)}-\frac{2}{s^2+2s+2} $
where $\dot{v_c}=j\omega v_c$

so now i need to break the fractures into a simpler ones
but here its all complex and i dont know how to get
a simpler fractures and there foorier transformation
??

so this is where i got stuck
and i even didnt get closer to the main solution of the problem

??

MrAl · 14th July 2009, 01:55 AM

Hi,

One way to look at the equations that have to be broken up as complex
is like this...
sin(wt) => w/(s^2+w)
and that has no damping so that makes sense right? because a sin wave
has no damping because it goes on forever.

On the other hand:
1/(s^2+s+1)
has an 's' term, so that has damping, so we should expect a form that
has e^at multiplied by some other oscillating term either sin or cos or both.

Another way to look at a form like:
1/(s^2+s+1)

is that the denominator has two solutions that form a complex pair:
a=(r+i*j) and b=(r-i*j)
and this means of course that:
1/(s^2+s+1)=1/(s-a)*1/(s-b)
so we break it apart just the same only using complex solutions.

Of course we could have used a table to look up q/(s^2+m*s+n) or something
like that, or we could use the fact that each part of the above
1/(s+A) and 1/(s+B)
both have transforms, and since we know they are of the form:
e^-At and e^-Bt
we could use those and the fact that multiplication in the frequency domain is
equivalent to convolution in the time domain. Thus, we can use the convolution
integral to develop a general formula for the complex convolution of
1/(s+A) and 1/(s+B)
and then use that formula to find the transform for this and other problems too.

Of course we could have looked this up in a table beforehand, but the solution to
the convolution integral above would be:

(e^-At-e^-Bt)/(B-A)

The only problem left now is that A and B are complex, so the result we end
up with is complex. To deal with this, we need the following relationships:

sin(wt)=-(j*(e^(j*wt)-e^(-j*wt)))/2
cos(wt)=(e^(j*wt)+e^(-j*wt))/2

which i believe are known as "Euler Identities" for sin and cos.

The solution comes in the form of an exponential damping factor multiplied
by a sin or cos factor. The sinusoidal part is best recognized by factoring
out the damping factor first, then equating to one of the above identities.

The solution for 1/(s^2+s+1) is:
(2*e^(-t/2)*sin((sqrt(3)*t)/2))/sqrt(3)

and note that the imaginary part of the solution of the denominator forms
the sin part and the real part forms the damping coefficient 'a' in e^at.

If rem right, when the denominator is normalized so that the highest power
of 's' is one, the exponential part is the coefficient of 's' divided by minus two, so
for:
1/(s^2+2*s+1)
(note the highest power of 's' has a coefficient of 1)
the exponential damping part would be:
e^(2/(-2)*t)=e^-t

transgalactic · 14th July 2009, 08:00 AM

$-\omega ^2V_c-2+2j\omega V_c +2V_c=cos(t)$

in the solution i have +2 instead of the -2

and why cos t turns to 1
??

MrAl · 14th July 2009, 08:56 AM

Quote:

Originally Posted by transgalactic

$-\omega ^2V_c-2+2j\omega V_c +2V_c=cos(t)$

in the solution i have +2 instead of the -2

and why cos t turns to 1
??

Hi again,

I see that you are attempting the solution first for vs=cos(t), and you
got pretty far, but unfortunately there is a mistake in the algebra
that you need to go back and check over again. Your intermediate
solution should come out to this:

(s^2+2*s+2)*Vs-2=s/(s^2+1)

and the only difference here is (and i think you noted) the minus
2 instead of +2, which makes all the difference.

As far as
"cos(t) turns to 1"
i can only guess that you mean
that when you work on part of the intermediate solution to
get the transform you end up with factors that lead to sin(t)
with no cos(t) present. I am guessing here that you want to
transform:
2/(s^2+2*s+2)
and you are coming up with a solution that includes sin(t) but
no cos(t). If this is the case, then you should again think about
the form as having no 's' in the numerator. When there is an 's'
in the numerator we end up with a cos term too.
Im not sure if that's what you meant though when you said that
the cos term goes to 1.

In any case, the transform for that part of the problem

2/(s^2+2*s+2)

is equal to:

2*e^(-t)*sin(t)

and there is clearly no cos term in there.

transgalactic · 14th July 2009, 10:50 AM

you did a Laplace tranformation of cosine
there is no s in phasors
i was wrong in putting s
i used in terms
$ s=j\omega $
so cosine need to be transformed by foorier
and
x^2+2x+2=0
x2=-1-i x1=-1+i
x1,2=alpha -+ wi
(w=omega)
omega is the coefficient of the Im part
$ \omega=1 $

transgalactic · 14th July 2009, 10:54 AM

the the cosine transforming to 1
is my problem

transgalactic · 14th July 2009, 10:00 PM

i found for the cosine input expression.
the phasor formula for v=Acos(xt+f) is v=Ae^{jf}
now i need to find the response for a step function
but i dont know its phasor formula

to what u(t) will turn to

??

how to transform the input of cos(t)+u(t)

?

transgalactic · 15th July 2009, 06:42 AM

sorry i forgot that i found already the response for step function

now i was told that the total answer is no
because of the starting conditions

what the problem with the starting positions so

the response for $v_s(t)=u(t)+cos(t)$
differs the sum of $v_c^u$ (the response for spet function) and $v_c^cos$ the response for cosine

MrAl · 15th July 2009, 09:19 AM

Hi,

I got the same result, that the sum of the individual responses is not the same
as the response with the two excitations at the same time. I also got the
same result that the non zero initial condition of v'(0)=2 caused the two
responses to be different and checked this by solving the problem twice,
once with v'(0)=2 and once with v'(0)=0.

First, with the initial conditions as stated v'(0)=2:

with vs=u(t) we get:
V=(2*s+1)/(s^3+2*s^2+2*s)

with vs=cos(t) we get:
V=(2*s^2+s+2)/(s^4+2*s^3+3*s^2+2*s+2)

and now suming the individual freq responses from u(t) and cos(t) we get:
Vsum=(4*s^3+2*s^2+4*s+1)/(s^5+2*s^4+3*s^3+2*s^2+2*s)

However, with vs=u(t)+cos(t) we get:
Vboth=(2*s^3+2*s^2+2*s+1)/(s^5+2*s^4+3*s^3+2*s^2+2*s)

Comparing the two Vsum to Vboth we can clearly see that they are not the same.

Second, lets set the initial condition of v'(0) to zero and look at this again:

and with u(t) we get:
1/(s^3+2*s^2+2*s)

and with cos(t) we get:
s/(s^4+2*s^3+3*s^2+2*s+2)

and adding these we get:
(2*s^2+1)/(s^5+2*s^4+3*s^3+2*s^2+2*s)

and with u(t)+cos(t) we get:
(2*s^2+1)/(s^5+2*s^4+3*s^3+2*s^2+2*s)

and now we can see that the added and the total responses are the same this time,
which shows that the initial condition of v'(0)=2 caused the difference.

It's also a little interesting to look at the two time responses, one for u(t)+cos(t) and the
other for the individual responses (all with v'(0)=2):

for vs=u(t)+cos(t) we get:
v(t)=e^(-t)*((9*sin(t))/10-(7*cos(t))/10)+(2*sin(t))/5+cos(t)/5+1/2

but for vs= individual u(t) and cos(t) added we get:
v(t)=e^(-t)*((29*sin(t))/10-(7*cos(t))/10)+(2*sin(t))/5+cos(t)/5+1/2

and note that the difference is in that one coefficient of sin(t), being 9 for one and 29 for the other.

transgalactic · 15th July 2009, 11:41 AM

what is the phasor transformation for this signal
u(t)+cos(t)

i lack only this small detail
i dont know how this transform this sum
??

transgalactic · 15th July 2009, 12:02 PM

for u(t) only response i did without phasors
i got
Vc=e^{-t}(-0.5cost+3sint)+0.5u(t)

for cos(t) using phasors i got
a long expression
and it differs ours too

please show the way
i cant understand how you got you result

can you scan the paper on which you have done these calculations
?

transgalactic · 15th July 2009, 01:02 PM

this is a formal explantion
"because the circuit is linear and time invariant then we can do the superposition here,
but the super position only works with ZSR and zsr is lenear regarding the components of the cercuit.
the superposition is not working for ZIR (which is linear for the starting conditions)."
i am a little weak with these theoretical terms:
ZSR is the private solution of a diff equation
ZIR is the homogenius solution of the diff equation

how does these gets fine if the starting conditions will be 0
why suddenly (in pure theoretic way) the super position will working in this case too
what will happen in the diff equation that will make the usage of the superposition priniciple be ok??

MrAl · 15th July 2009, 11:29 PM

Hi again,

You have two questions, one has a short answer, the other long...

When you ask to transform the source:

u(t)+cos(t)

into 'phasor' form this doesnt make sense because u(t) is not considered
an AC signal, and phasors are for AC analysis. cos(t) does have a phasor
form and that is amplitude separated by a little 'angle' symbol followed by
0 degrees similar to this:
1_0
where 1 is amplitude and 0 is the phase angle, only the underscore is actually
the little 'angle' symbol usually used for phasors.
See the attached drawing for phasor representation.
Did you instead mean you wanted the Laplace or Fourier transform?

Ok, back to the differential equation and how it is different with initial condition
of v'(0)=2 than with v'(0)=0 (which changes the basic type of differential equation)...

These are the Laplace forms for the function and its derivatives:

Code:

FORM     ALT FORM    FORMAL s FORM
f(t)     f(t)        F(s)
f'(t)    df/dt       s*F(s)-f(0)
f''(t)   d^2f/dt^2   s^2*F(s)-s*f(0)-f'(0)

Example of using the Laplace derivative forms to solve the differential equation:

v''+2*v'+2*v=vs(t) with initial conditions: v'(0)=2 and v(0)=0

Find out if the response to vs(t)=u(t)+cos(t) is the same as the sum of the individual responses
to u(t) and cos(t) alone. Later we want to find out how the solution is different when we
change the initial condition for v'(0) from non zero to zero.

Here we need to solve the differential equation three times, first with vs(t)=u(t), then with
vs(t)=cos(t) and then sum the two, and finally with vs(t)=u(t)+cos(t), then compare these
two results to see if they are the same.

We will use the following notation to make the writing simpler:

Code:

  v''  =vdd
  v'   =vd
  v    =v
  v'(0)=vd0
  v(0) =v0
  vs(t)=vs
  V(s) =V

This makes the differential equation look like this:
vdd+2*vd+2*v=vs

Transforming the various parts:
v=V
vd=s*V-v0
vdd=s^2*V-s*v0-vd0

Substituting into the diff equ:
s^2*V-s*v0-vd0+2*(s*V-v0)+2*V=vs
and simplifying:
s^2*V-s*v0-vd0+2*s*V-2*v0+2*V=vs
and grouping all the V terms:
V*(s^2+2*s+2)-s*v0-vd0-2*v0=vs
and since v0=0 this simplifies to:
V*(s^2+2*s+2)-vd0=vs
and since vd0=2 this simplifies to:
V*(s^2+2*s+2)-2=vs
or:
V*(s^2+2*s+2)=vs+2
or:
V=(vs+2)/(s^2+2*s+2)
and now we have a generalized form we can use to look at the three possible sources for vs.

The three sources and their transforms are:
vs=u(t) => 1/s
vs=cos(t) => s/(s^2+1)
vs=u(t)+cos(t) => 1/s+s/(s^2+1)

Using the generalized form from above:
V=(vs+2)/(s^2+2*s+2)

All we have to do now is substitute one of the sources for vs and then carry out the
algebraic simplification. For example, with vs=u(t) the transform is 1/s, so we get:
V=(1/s+2)/(s^2+2*s+2)
and multiplying top and bottom by s we get:
V=(2*s+1)/(s^3+2*s^2+2*s)
This is all we really have to do because we can later add this to the response from
cos(t) and then compare that to the response from both sources together. We dont
really need the time response to do this.

And now the responses to these three inputs (we dont really need the time equations here though)

with vs=u(t) we get:
V=(2*s+1)/(s^3+2*s^2+2*s)
v(t)=e^(-t)*((3*sin(t))/2-cos(t)/2)+1/2

with vs=cos(t) we get:
V=(2*s^2+s+2)/(s^4+2*s^3+3*s^2+2*s+2)
v(t)=e^(-t)*((7*sin(t))/5-cos(t)/5)+(2*sin(t))/5+cos(t)/5

Suming the individual freq responses from u(t) and cos(t) we get:
V=(4*s^3+2*s^2+4*s+1)/(s^5+2*s^4+3*s^3+2*s^2+2*s)
v(t)=e^(-t)*((29*sin(t))/10-(7*cos(t))/10)+(2*sin(t))/5+cos(t)/5+1/2

With vs=u(t)+cos(t) we get:
V=(2*s^3+2*s^2+2*s+1)/(s^5+2*s^4+3*s^3+2*s^2+2*s)
v(t)=e^(-t)*((9*sin(t))/10-(7*cos(t))/10)+(2*sin(t))/5+cos(t)/5+1/2

Comparing the sum of individual responses to the response of both sources at the same time
shows that the two responses are not the same. Note that we only have to compare:

V=(4*s^3+2*s^2+4*s+1)/(s^5+2*s^4+3*s^3+2*s^2+2*s)

to

V=(2*s^3+2*s^2+2*s+1)/(s^5+2*s^4+3*s^3+2*s^2+2*s)

to see that they are different, we dont have to compare the time responses.

Now we want to see what changes when we change the initial condition of v'(0) to zero
instead of 2:

Starting with the Laplace form of the diff equ:
s^2*V-s*v0-vd0+2*(s*V-v0)+2*V=vs
and simplifying:
s^2*V-s*v0-vd0+2*s*V-2*v0+2*V=vs
and grouping all the V terms:
V*(s^2+2*s+2)-s*v0-vd0-2*v0=vs
and since v0=0 this simplifies to:
V*(s^2+2*s+2)-vd0=vs
but now vd0=0, not 2 like before, so:
vd0=0
and since vd0=0 now this simplifies to:
V*(s^2+2*s+2)=vs
and note the previous form with vd0=2 was V*(s^2+2*s+2)-2=vs
so we lost the "-2" when vd0 went to zero.

Simplifying:
V*(s^2+2*s+2)=vs
or:
V=vs/(s^2+2*s+2)
and now we have a generalized form we can use to look at the three possible sources for vs
again. Note however that the numerator does not contain that constant term anymore, so
the solution with u(t)+cos(t) comes out the same as the solution with added individual
responses to u(t) and cos(t). I'll leave it up to you to do the rest of the math to show this.

Lastly, here is a drawing showing the phasor representation of an AC signal:

transgalactic · 16th July 2009, 07:35 PM

thanks

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13th July 2009, 04:44 PM	#1
transgalactic	using phazors question.. this is the last part of a bigger question if you i forgot to mention some data please tell in the last part i got $\ddot{v_c}+2\dot{v_c}+2v_c=v_s(t)$ $\dot{v_c(0)}=2$ $v_c(0)=0$ i need to find out if the response for $v_s(t)=u(t)+cos(t)$ equals the sum of $v_c^u$ (the response for spet function) and $v_c^cos$ the response for cosine i start by finding the response for cos(t) $v_c(t)->V_c(s)$ $\dot{v_c}->sV_c(s)-v_c(0)$ $\ddot{v_c}->s(sV_c(s)-v_c(0))-\dot{v_c(0)}$ $s(sV_c(s)-v_c(0))-\dot{v_c(0)}+2sV_c(s)-v_c(0)+2V_c(s)=\frac{s}{s^2+1}$ $<br /> V_c(s)[s^2+2s+2]+2=\frac{s}{s^2+1}<br />$ so i get $<br /> V_c(s)=\frac{s}{(s^2+2s+2)(s^2+1)}-\frac{2}{s^2+2s+2}<br />$ where $\dot{v_c}=j\omega v_c$ so now i need to break the fractures into a simpler ones but here its all complex and i dont know how to get a simpler fractures and there foorier transformation ?? so this is where i got stuck and i even didnt get closer to the main solution of the problem ?? Last edited by transgalactic; 13th July 2009 at 05:07 PM.

14th July 2009, 01:55 AM	#2
MrAl	Hi, One way to look at the equations that have to be broken up as complex is like this... sin(wt) => w/(s^2+w) and that has no damping so that makes sense right? because a sin wave has no damping because it goes on forever. On the other hand: 1/(s^2+s+1) has an 's' term, so that has damping, so we should expect a form that has e^at multiplied by some other oscillating term either sin or cos or both. Another way to look at a form like: 1/(s^2+s+1) is that the denominator has two solutions that form a complex pair: a=(r+ij) and b=(r-ij) and this means of course that: 1/(s^2+s+1)=1/(s-a)1/(s-b) so we break it apart just the same only using complex solutions. Of course we could have used a table to look up q/(s^2+ms+n) or something like that, or we could use the fact that each part of the above 1/(s+A) and 1/(s+B) both have transforms, and since we know they are of the form: e^-At and e^-Bt we could use those and the fact that multiplication in the frequency domain is equivalent to convolution in the time domain. Thus, we can use the convolution integral to develop a general formula for the complex convolution of 1/(s+A) and 1/(s+B) and then use that formula to find the transform for this and other problems too. Of course we could have looked this up in a table beforehand, but the solution to the convolution integral above would be: (e^-At-e^-Bt)/(B-A) The only problem left now is that A and B are complex, so the result we end up with is complex. To deal with this, we need the following relationships: sin(wt)=-(j(e^(jwt)-e^(-jwt)))/2 cos(wt)=(e^(jwt)+e^(-jwt))/2 which i believe are known as "Euler Identities" for sin and cos. The solution comes in the form of an exponential damping factor multiplied by a sin or cos factor. The sinusoidal part is best recognized by factoring out the damping factor first, then equating to one of the above identities. The solution for 1/(s^2+s+1) is: (2e^(-t/2)sin((sqrt(3)t)/2))/sqrt(3) and note that the imaginary part of the solution of the denominator forms the sin part and the real part forms the damping coefficient 'a' in e^at. If rem right, when the denominator is normalized so that the highest power of 's' is one, the exponential part is the coefficient of 's' divided by minus two, so for: 1/(s^2+2s+1) (note the highest power of 's' has a coefficient of 1) the exponential damping part would be: e^(2/(-2)t)=e^-t Last edited by MrAl; 14th July 2009 at 02:10 AM.

14th July 2009, 08:00 AM	#3
transgalactic	$-\omega ^2V_c-2+2j\omega V_c +2V_c=cos(t)$ in the solution i have +2 instead of the -2 and why cos t turns to 1 ?? Last edited by transgalactic; 14th July 2009 at 08:23 AM.

14th July 2009, 10:50 AM	#5
transgalactic	you did a Laplace tranformation of cosine there is no s in phasors i was wrong in putting s i used in terms $<br /> s=j\omega <br />$ so cosine need to be transformed by foorier and x^2+2x+2=0 x2=-1-i x1=-1+i x1,2=alpha -+ wi (w=omega) omega is the coefficient of the Im part $<br /> \omega=1<br />$

14th July 2009, 10:54 AM	#6
transgalactic	the the cosine transforming to 1 is my problem

14th July 2009, 10:00 PM	#7
transgalactic	i found for the cosine input expression. the phasor formula for v=Acos(xt+f) is v=Ae^{jf} now i need to find the response for a step function but i dont know its phasor formula to what u(t) will turn to ?? how to transform the input of cos(t)+u(t) ? Last edited by transgalactic; 14th July 2009 at 10:15 PM.

15th July 2009, 06:42 AM	#8
transgalactic	sorry i forgot that i found already the response for step function now i was told that the total answer is no because of the starting conditions what the problem with the starting positions so the response for $v_s(t)=u(t)+cos(t)$ differs the sum of $v_c^u$ (the response for spet function) and $v_c^cos$ the response for cosine

15th July 2009, 09:19 AM	#9
MrAl	Hi, I got the same result, that the sum of the individual responses is not the same as the response with the two excitations at the same time. I also got the same result that the non zero initial condition of v'(0)=2 caused the two responses to be different and checked this by solving the problem twice, once with v'(0)=2 and once with v'(0)=0. First, with the initial conditions as stated v'(0)=2: with vs=u(t) we get: V=(2s+1)/(s^3+2s^2+2s) with vs=cos(t) we get: V=(2s^2+s+2)/(s^4+2s^3+3s^2+2s+2) and now suming the individual freq responses from u(t) and cos(t) we get: Vsum=(4s^3+2s^2+4s+1)/(s^5+2s^4+3s^3+2s^2+2s) However, with vs=u(t)+cos(t) we get: Vboth=(2s^3+2s^2+2s+1)/(s^5+2s^4+3s^3+2s^2+2s) Comparing the two Vsum to Vboth we can clearly see that they are not the same. Second, lets set the initial condition of v'(0) to zero and look at this again: and with u(t) we get: 1/(s^3+2s^2+2s) and with cos(t) we get: s/(s^4+2s^3+3s^2+2s+2) and adding these we get: (2s^2+1)/(s^5+2s^4+3s^3+2s^2+2s) and with u(t)+cos(t) we get: (2s^2+1)/(s^5+2s^4+3s^3+2s^2+2s) and now we can see that the added and the total responses are the same this time, which shows that the initial condition of v'(0)=2 caused the difference. It's also a little interesting to look at the two time responses, one for u(t)+cos(t) and the other for the individual responses (all with v'(0)=2): for vs=u(t)+cos(t) we get: v(t)=e^(-t)((9sin(t))/10-(7cos(t))/10)+(2sin(t))/5+cos(t)/5+1/2 but for vs= individual u(t) and cos(t) added we get: v(t)=e^(-t)((29sin(t))/10-(7cos(t))/10)+(2sin(t))/5+cos(t)/5+1/2 and note that the difference is in that one coefficient of sin(t), being 9 for one and 29 for the other. Last edited by MrAl; 15th July 2009 at 09:26 AM.

15th July 2009, 11:41 AM	#10
transgalactic	what is the phasor transformation for this signal u(t)+cos(t) i lack only this small detail i dont know how this transform this sum ??

15th July 2009, 12:02 PM	#11
transgalactic	for u(t) only response i did without phasors i got Vc=e^{-t}(-0.5cost+3sint)+0.5u(t) for cos(t) using phasors i got a long expression and it differs ours too please show the way i cant understand how you got you result can you scan the paper on which you have done these calculations ? Last edited by transgalactic; 15th July 2009 at 12:10 PM.

15th July 2009, 01:02 PM	#12
transgalactic	this is a formal explantion "because the circuit is linear and time invariant then we can do the superposition here, but the super position only works with ZSR and zsr is lenear regarding the components of the cercuit. the superposition is not working for ZIR (which is linear for the starting conditions)." i am a little weak with these theoretical terms: ZSR is the private solution of a diff equation ZIR is the homogenius solution of the diff equation how does these gets fine if the starting conditions will be 0 why suddenly (in pure theoretic way) the super position will working in this case too what will happen in the diff equation that will make the usage of the superposition priniciple be ok?? Last edited by transgalactic; 15th July 2009 at 01:03 PM.

15th July 2009, 11:29 PM	#13
MrAl	Hi again, You have two questions, one has a short answer, the other long... When you ask to transform the source: u(t)+cos(t) into 'phasor' form this doesnt make sense because u(t) is not considered an AC signal, and phasors are for AC analysis. cos(t) does have a phasor form and that is amplitude separated by a little 'angle' symbol followed by 0 degrees similar to this: 1_0 where 1 is amplitude and 0 is the phase angle, only the underscore is actually the little 'angle' symbol usually used for phasors. See the attached drawing for phasor representation. Did you instead mean you wanted the Laplace or Fourier transform? Ok, back to the differential equation and how it is different with initial condition of v'(0)=2 than with v'(0)=0 (which changes the basic type of differential equation)... These are the Laplace forms for the function and its derivatives: Code: FORM ALT FORM FORMAL s FORM f(t) f(t) F(s) f'(t) df/dt sF(s)-f(0) f''(t) d^2f/dt^2 s^2F(s)-sf(0)-f'(0) Example of using the Laplace derivative forms to solve the differential equation: v''+2v'+2v=vs(t) with initial conditions: v'(0)=2 and v(0)=0 Find out if the response to vs(t)=u(t)+cos(t) is the same as the sum of the individual responses to u(t) and cos(t) alone. Later we want to find out how the solution is different when we change the initial condition for v'(0) from non zero to zero. Here we need to solve the differential equation three times, first with vs(t)=u(t), then with vs(t)=cos(t) and then sum the two, and finally with vs(t)=u(t)+cos(t), then compare these two results to see if they are the same. We will use the following notation to make the writing simpler: Code: v'' =vdd v' =vd v =v v'(0)=vd0 v(0) =v0 vs(t)=vs V(s) =V This makes the differential equation look like this: vdd+2vd+2v=vs Transforming the various parts: v=V vd=sV-v0 vdd=s^2V-sv0-vd0 Substituting into the diff equ: s^2V-sv0-vd0+2(sV-v0)+2V=vs and simplifying: s^2V-sv0-vd0+2sV-2v0+2V=vs and grouping all the V terms: V(s^2+2s+2)-sv0-vd0-2v0=vs and since v0=0 this simplifies to: V(s^2+2s+2)-vd0=vs and since vd0=2 this simplifies to: V(s^2+2s+2)-2=vs or: V(s^2+2s+2)=vs+2 or: V=(vs+2)/(s^2+2s+2) and now we have a generalized form we can use to look at the three possible sources for vs. The three sources and their transforms are: vs=u(t) => 1/s vs=cos(t) => s/(s^2+1) vs=u(t)+cos(t) => 1/s+s/(s^2+1) Using the generalized form from above: V=(vs+2)/(s^2+2s+2) All we have to do now is substitute one of the sources for vs and then carry out the algebraic simplification. For example, with vs=u(t) the transform is 1/s, so we get: V=(1/s+2)/(s^2+2s+2) and multiplying top and bottom by s we get: V=(2s+1)/(s^3+2s^2+2s) This is all we really have to do because we can later add this to the response from cos(t) and then compare that to the response from both sources together. We dont really need the time response to do this. And now the responses to these three inputs (we dont really need the time equations here though) with vs=u(t) we get: V=(2s+1)/(s^3+2s^2+2s) v(t)=e^(-t)((3sin(t))/2-cos(t)/2)+1/2 with vs=cos(t) we get: V=(2s^2+s+2)/(s^4+2s^3+3s^2+2s+2) v(t)=e^(-t)((7sin(t))/5-cos(t)/5)+(2sin(t))/5+cos(t)/5 Suming the individual freq responses from u(t) and cos(t) we get: V=(4s^3+2s^2+4s+1)/(s^5+2s^4+3s^3+2s^2+2s) v(t)=e^(-t)((29sin(t))/10-(7cos(t))/10)+(2sin(t))/5+cos(t)/5+1/2 With vs=u(t)+cos(t) we get: V=(2s^3+2s^2+2s+1)/(s^5+2s^4+3s^3+2s^2+2s) v(t)=e^(-t)((9sin(t))/10-(7cos(t))/10)+(2sin(t))/5+cos(t)/5+1/2 Comparing the sum of individual responses to the response of both sources at the same time shows that the two responses are not the same. Note that we only have to compare: V=(4s^3+2s^2+4s+1)/(s^5+2s^4+3s^3+2s^2+2s) to V=(2s^3+2s^2+2s+1)/(s^5+2s^4+3s^3+2s^2+2s) to see that they are different, we dont have to compare the time responses. Now we want to see what changes when we change the initial condition of v'(0) to zero instead of 2: Starting with the Laplace form of the diff equ: s^2V-sv0-vd0+2(sV-v0)+2V=vs and simplifying: s^2V-sv0-vd0+2sV-2v0+2V=vs and grouping all the V terms: V(s^2+2s+2)-sv0-vd0-2v0=vs and since v0=0 this simplifies to: V(s^2+2s+2)-vd0=vs but now vd0=0, not 2 like before, so: vd0=0 and since vd0=0 now this simplifies to: V(s^2+2s+2)=vs and note the previous form with vd0=2 was V(s^2+2s+2)-2=vs so we lost the "-2" when vd0 went to zero. Simplifying: V(s^2+2s+2)=vs or: V=vs/(s^2+2s+2) and now we have a generalized form we can use to look at the three possible sources for vs again. Note however that the numerator does not contain that constant term anymore, so the solution with u(t)+cos(t) comes out the same as the solution with added individual responses to u(t) and cos(t). I'll leave it up to you to do the rest of the math to show this. Lastly, here is a drawing showing the phasor representation of an AC signal: Attached Thumbnails Last edited by MrAl; 15th July 2009 at 11:55 PM.*